Question

Find the sum of first 45 term of an A.p whose 8 term is 41 and 13 term is 61​

Answer 1

Step-by-step explanation:

$given \\ \: t8 = 41 \\ and \\ t13 = 61$

$we \: know \: that \: \\ tn = a + (n - 1)d$

$so \: applying \: same \: formula$

$t8 = a + 7d \\ t13 = a + 12d$

$41 = a + 7d \\ 61 = a + 12d$

$now \: lets \: solve \: this \: equations$

$5d = 20 \\ d = 4$

$now \: substitute \: d = 4 \: in \: one \: of \: the \: equations$

$41 = a + 7d \\ 41 = a + 28 \\ a = 13$

$we \: know \: the \: formula \: for \: sn \: = \frac{n}{2} {2a + (n - 1)d}$

$s41 = \frac{41}{2} ({26 + 160}) \\ \: \: \: \: \: \: \: \: = \frac{41}{2} (186) \\ \: \: \: \: \: \: \: \: = 41 \times 93 \\ \: \: \: \: \: \: \: \: = 3813$

$hey \: now \: we \: have \: done \: with \: this \: question \: hope \: it \: helps \: uh$